从嵌套字典中删除键（Python 键）答案

【问题标题】：Remove keys from a nested dict (Python keys)从嵌套字典中删除键（Python 键）
【发布时间】：2019-11-18 21:56:57
【问题描述】：

我是 Python 新手，提前感谢您的帮助。

我构建了以下代码（我尝试了以下代码，我在字典中使用了字典）。

这个想法是保持键（hair.color）和值（金发）。在此示例中：删除 Micheal。

代码：

def answers(hair_questions):
    try:
        for i in people:
            if people[i]["hair.color"]==hair_questions:
                print(people[i])
            else:
                del people[i]
            return people[i]
    except:
        print("Doesn´t exist")

answers("brown")

关于人：

people={
 "Anne":
    {
   "gender":"female",
   "skin.color":"white",
  "hair.color":"blonde",
  "hair.shape":"curly"
 }
,
"Michael":
{

  "citizenship":"africa",
  "gender":"male",
  "hair.color":"brown",
  "hair.shape":"curly"


}
,

"Ashley":
    {
  "gender":"female",
  "citizenship":"american",
  "hair.color":"blonde",
  "hair.shape":"curly "
 }

 }

代码仅检查第一个键：在条件下：values(blonde) 即(people[i]["hair.color"]!=brown) 它仅适用于 1 个键，然后代码“卡住”

我目前的输出：

"people"=

 "Michael":
{

  "citizenship":"africa",
  "gender":"male",
  "hair.color":"brown",
  "hair.shape":"curly"


}
,

"Ashley":
    {
  "gender":"female",
  "citizenship":"american",
  "hair.color":"blonde",
  "hair.shape":"curly "
 }

相反，我想要：

"people"=

"Michael":
{

  "citizenship":"africa",
  "gender":"male",
  "hair.color":"brown",
  "hair.shape":"curly"

}

我想要一个输出，在这种情况下，（仅）Michael。

【问题讨论】：

您是否严格需要删除键，或者创建一个只包含所需键的新字典就足够了？
丢掉 else 部分，你就可以开始了
如果你只想要hair.color == brown的那些，我也不确定为什么Michael会出现在输出中

标签： python string dictionary

【解决方案1】：

在 for 循环迭代时不能删除键：

people={
    "Anne":
        {
       "gender":"female",
       "skin.color":"white",
      "hair.color":"blonde",
      "hair.shape":"curly"
     },
    "Michael":
    {
      "citizenship":"africa",
      "gender":"male",
      "hair.color":"brown",
      "hair.shape":"curly"
    },
    "Ashley":
        {
          "gender":"female",
          "citizenship":"american",
          "hair.color":"blonde",
          "hair.shape":"curly "
        }
 }

def answers(hair_questions):
    my_dict = {}
    for i in people:
        if people[i]["hair.color"] in hair_questions:
            my_dict[i] = people[i]
    return  my_dict

print(answers("brown"))

或

def answers(hair_questions):
    my_list = []
    for i in people:
        if people[i]["hair.color"] not in hair_questions:
            my_list.append(i)

    for i in my_list:
        del people[i]

answers("brown")
print(people)

O/P：

{'Michael': {'citizenship': 'africa', 'gender': 'male', 'hair.color': 'brown', 'hair.shape': 'curly'}}

【讨论】：

【解决方案2】：

您可以使用列表推导：

brown = {key:value for key,value in people.items() if people[key]["hair.color"] != "blonde"}
print (brown)

等于：

brown= {}
for key,value in people.items():
    if people[key]["hair.color"] != "blonde":
        brown[key] = value
print (brown)

输出：

{'Michael': {'citizenship': 'africa', 'gender': 'male', 'hair.color': 'brown', 'hair.shape': 'curly'}}

【讨论】：